Armstrong FM modulator: find n1 and n2

Given: audio 50 Hz – 15 kHz; $f_{c1} = 0.2$ MHz; $\beta_1 = 0.5$; $f_{c2} = 10.925$ MHz; final $f_c = 90$ MHz, $\Delta f = 75$ kHz.

 NB PM     x n1        Mixer          x n2
 0.2 MHz->[n1]->0.2n1 MHz->[X]->f2->[n2]-> 90 MHz
 df1=25 Hz                  ^             df=75 kHz
                     [10.925 MHz]

Step 1: deviation at the NB modulator

The NB phase modulator (with integrator) must keep $\beta_1 \le 0.5$ for every audio frequency. Since $\beta_1 = \Delta f_1/f_m$, the worst case is the lowest audio frequency, 50 Hz:

$$\Delta f_1 = \beta_1f_{m,min} = 0.5 \times 50 = 25\ \text{Hz}$$

Step 2: total multiplication needed

The mixer does not change the deviation, so

$$n_1n_2 = \frac{\Delta f}{\Delta f_1} = \frac{75 \times 10^3}{25} = 3000$$

Step 3: carrier frequency condition

The mixer output is the difference frequency (taking $n_1f_{c1} > f_{c2}$):

$$f_c = n_2(n_1f_{c1} - f_{c2}) \;\Rightarrow\; 90 = n_2(0.2n_1 - 10.925)$$

Using $n_1n_2 = 3000$: $0.2n_1n_2 = 600$, so

$$\begin{aligned} 90 &= 600 - 10.925n_2 \\ n_2 &= \frac{510}{10.925} = 46.68, \qquad n_1 = \frac{3000}{46.68} = 64.27 \end{aligned}$$

Multipliers must be integers. Choosing $n_1 = 64$ and solving the frequency equation exactly:

$$\begin{aligned} 0.2 \times 64 &= 12.8\ \text{MHz} \\ f_2 &= 12.8 - 10.925 = 1.875\ \text{MHz} \\ n_2 &= \frac{90}{1.875} = 48 \end{aligned}$$

Check

$$\begin{aligned} f_c &= 48 \times 1.875 = 90\ \text{MHz} \\ \Delta f &= n_1n_2\Delta f_1 = 64 \times 48 \times 25 = 76800\ \text{Hz} = 76.8\ \text{kHz} \end{aligned}$$

The carrier is exactly 90 MHz and the deviation is within 2.4% of 75 kHz (exactly 75 kHz is reached by reducing $\beta_1$ slightly to 0.488). Both values are practical: $64 = 2^6$ (six doublers) and $48 = 2^4 \times 3$ (four doublers and a tripler).

Answer: $n_1 = 64$, $n_2 = 48$ (ideal product 3000; continuous solution $n_1 \approx 64.3$, $n_2 \approx 46.7$).